shm review solutions

8/3/2019 SHM Review Solutions

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PC4110 Advanced Physics I SolutionsSimple Harmonic Motion

Answers - Simple Harmonic Motion

1.

a. The maximum displacement from the equilibrium position A =10.0 cm.

b. The time for one complete oscillation T = p/2 s. Notice themaximum positive displacement x = +10.0 cm occurs at t =0 and the next time at t = p/2 s. It occurs again at t = ps.

2.

a. v(t) = - (2p/T)(A sin 2pt/T). The maximum value of the sine is1. The maximum absolute value of v = 2pA/T. The signsaccount only for the direction of the velocity.

b. From Fig. 1b above, |vmax| = 40.0 cm/s = 2pA/T =

2p(10cm)/ps/2.

3.

2012 Semester 1 Term 1 1


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a.

i. a(t) =- (2p/T)2(A cos 2pt/T). Maximum value of thecosine is1. |amax| = (2p/T)2A. The signs account only for thedirection of the acceleration.

ii. a(t) = - (2p/T)2 (A cos 2pt/T) = - (2p/T)2 x(t)since x(t) = A cos 2pt/T.

-a(t)/x(t) = (2p/T)2.

b. From Fig. 1c above,

|amax| = 160.0 cm/s2 = (2p/T)2A = (2p/ps/2)2 10 cm.

4.a.

For a mass attached to a spring, F = - kx or F/x = - k, where kis a constant. The applied force F is directly proportional to thedisplacement x and the minus sign says it is in the oppositedirection to x.

b. When the spring is extended and released, Fnet = ma or- kx = ma or - a/x = k/m.

c. From 3a(ii), - a(t)/x(t) = (2p/T)2

By comparison, k/m = (2p/T)2 or T = 2p(m/k)1/2.

5. a. Given x(t) = A cos (2pt/T + d), where A is the maximumdisplacement from the equilibrium position. The maximumvalue of cos (2pt/T + d) is 1, so the equation accuratelydescribe the definition of A.

b. x(t + T) = A cos [2p(t + T)/T + d]= A cos [2pt/T + 2pT/T+ d]= A cos [2pt/T + 2p + d]



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= A cos (2pt/T + d)= x(t).

The definition of T is accurately described by the equation of

motion for simple harmonic motion, A cos (2pt/T + d),because it allows the value of x at t to equal the value of xat t + T or t + nT where n = 1, 2, 3 . . .

c. For x(t) = A cos (2pt/T -p /2), x(0) = xo = A cos (- p /2) = 0.v(t) = - (2p/T)A sin (2pt/T- p /2).v(0) = vo = - (2p/T)A sin (- p /2) = (2p/T)A = +vmax.At t = 0, the object is at the equilibrium position andtravelling with the maximum velocity in the +X-direction.

6.Given that d2x/dt2 + (k/m)x = 0 (Equation

1")Show that x(t) = A cos (2p t/T + d) is a solution. (Equation1')v = dx/dt = - (2p/T)A sin (2pt/T + d).dv/dt = d2x/dt2 = - (2p/T)2[A cos (2p\t/T + d)] = - (2p/T)2 x

(Equation 2)

Substituting Eq. 2 into Eq. 1", - (2p/T)2 x + (k/m)x = 0.This equation is true if (2p/T)2 = (k/m) or T = 2p(m/k)1/2.

7.

In general, x(t) = A cos (2pt/T + d) and v(t) = dx/dt = -A(2p/T) sin(2pt/T + d).



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(a) and (d)

In Fig. for #7a above, x(t) is plotted for d = 0.

In Fig. for #7d above, v(t) is plotted for d = 0.For d = 0 and t = 0 , the initial conditions are:

the initial displacement = xo = A andthe initial velocity = vo = 0.

Immediately after t = 0, the object moves to the left (with anegative velocity).

(b) For d = - p/2, x(t) = A cos (2pt/T - p/2).

By trigonometric identity,cos (C - D) = cos C cos D + sin C sin D (with C = 2pt/T and D = p/2)x(t) = A[cos 2pt/T cos p/2 + sin 2pt/T sin p/2]



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x(t) = A[cos 2pt/T(0) + sin 2pt/T(1)] = A sin 2pt/T

Notice the first maximum in Fig. b (immediately above) lags that inFig. a (above) by p/2 radians.

(e) v(t) = - 2pA/T sin (2pt/T - p/2)

By trigonometric identity,sin (C - D) = sin C cos D - cos C sin Dv(t) = -2pA/T[(sin 2pt/T)(0) - (cos2pt/T)(1)]v(t) = (2pA/T) cos 2pt/T

For d = - p/2 and t = 0, xo = 0 and vo = +2pA/T

(c) For d = - p, x(t) = A cos (2pt/T - p)= A[cos 2pt/T cos p + sin 2pt/T sin p]

= A[(cos 2pt/T)(-1) + (sin 2pt/T)(0)]x(t) = - A cos 2pt/T

Notice that Fig. c (immediately above) lags Fig. a (above) by pradians.

(f) v(t) = - A(2p/T) sin (2pt/T - p)= - (2pA/T)[sin 2pt/T cos p-cos 2pt/T sin p]=(- 2pA/T)[(sin2pt/T)(-1) - (cos2pt/T)(0)].

v(t) = (2pA/T) sin 2pt/T

For d = - p and t = 0, xo = - A and vo = 0.

Immediately after t = 0, the object moves to the right with apositive velocity.

(g) The function of d is to state the initial conditions.Note: d = p or - p gives the same result.

8. a. x(t) = A cos (wt + d)x(0) = xo = A cos d (Equation 1)

b. v(t) = dx/dt = - wA sin (wt + d)v(0) = vo = -wA sin d (Equation 2)

Dividing both sides of Eq. 2 by - w:- vo/w = A sin d (Equation 3)

c. Dividing Eq. 3 by Eq. 1, tan d = - vo/xwo.



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d. Squaring Eq. 1 and Eq. 3 and adding:xo2 + (- vo/w)2 = A2(cos2 d + A sin2 d) or [xo2 + (- vo/w)2]1/2 = A.

9.

In Fig. for #9(a) above, the spring is not stretched. In Fig. for #9(b)above, the mass is attached and the spring is stretched a distancexo.

a. The mass comes to rest and Fnet = ma = m(0).Taking down to be positive,

- kxo + mg = 0 (Equation 1)

b. In Fig. for #9(c) above, the spring has been displaced anadditional distance x. Now Fnet = ma, where a 0 once the

spring is released. Taking the direction of x, which is down, aspositive,

- kx - kxo + mg = ma (Equation 2)

From Eq. 1, we see that - kxo + mg = 0. Eq. 2 becomes:

- kx = ma or - a/x = k/m.

The ratio of a to x is the same whether the spring ismounted horizontally or vertically.

c. As before, - a/x = (2p/T)2 = (2pf)2 = k/m and f = (1/2p)(k/m)1/2.

10.



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a. The forces acting on the pendulum bob are its weight mg andthe tension T in the string.

b. The only force tangent to the path is a restoring force - mg sinQ. From the triangle with lengths, we find that

sin Q = x/L and - mg sin Q = - (mg/L)x.

For small displacements, x s, we can think of thedisplacement and the restoring force acting horizontally.

Fnet = ma- (mg/L)x = ma

c. Since m, g, and L are constants, the restoring force, - (mg/L)x,

is directly proportional to the displacement and in the oppositedirection. The pendulum is an example of simple harmonicmotion.-a/x = (g/L) = (2p/T)2 = (2pf)2. f = (1/2p)(g/L)1/2.

For another approach, write Q = s/L. For small angles Q isapproximately equal to sin Q. The restoring force - mg sin Q - mgQ = m d2s/dt2

= m d2(LQ)/dt2, or - (g/L)Q = d2Q/dt2 and d2Q/dt2 + (g/L)Q = 0.This is the same form as d2x/dt2 + (k/m)x = 0, for which T =2p (m/k)1/2 and x = A cos (2p/T). By comparison with thespring, for the pendulum T = 2p (L/g)1/2 and f = 1/T = (1/2p)(g/L)1/2 and Q = Qmax cos (2p/T).

11Given x(t) = 0.01 m cos (0.02p s-1 t - p/2) compare with x(t) = A cos



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.(2p t/T + d) and find:

a. the amplitude A = 0.01 m

b. 2p/T = 0.02p s-1; period T = 100 s,

c. the frequency f = 1/T = 0.01 s-1, and

d. the initial phase d = - p/2.

12.

From Fig. 3 above, we see that:

a. The cosine curve repeats itself every 4.0s so the period T =4.0 s.

b. The amplitude of the motion A =10.0 cm.

c. If we write the equation of motion as a function of the cosine,

we let d = 0.x(t) = A cos 2pt/T = 10.0 cm cos 2pt/4.0 s = 10.0 cm cos pt s-1/2.

d. v(t) = dx/dt = -10.0p/2 cm/s sin pt s-1/2|vmax| = 5.0p cm/s

e. a(t) = dv/dt = -10.0 (p/2)2cm/s2 cos pt s-1/2|amax| = 2.5p2 cm/s2

13



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.

For the X-component:

(a) x = r cos Q = r cos wt

(b) vx = dx/dt = - wr sin wt

ax = dvx/dt = - w

2

(r cos wt) = - w

2

x

(c) ax/x = - w2 = - 4p2f2

(d) (e) Since the acceleration is directly proportional to thedisplacement and in the opposite direction, the motion is simpleharmonic. Remember by Newton's second law of motion theacceleration is directly proportional to, and in the same direction as,the force.

(f) ax/x = - w2 = - 4p2f2 = - k/m or w = 2pf = (k/m)1/2

For the Y-component:

(a) y = r sin Q = r sin wt

(b) vy = dy/dt = wr cos wtay = dvy/dt = - w2(r sin wt) = - w2y

(c) ay/y = - w2 = - 4p2f2

(d) (e) The motion is again simple harmonic, and

(f) ay/y = - w2

= - 4p2

f2

= - k/m or w = 2pf = (k/m)1/2

14.For a mass-spring system,

- kx = ma = md2x/dt2 ord2x/dt2 + (k/m)x = 0, where the period T = 2p (m/k)1/2.



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By comparison withb2d2x/dt2 + c2x = 0 ord2x/dt2 + (c/b)2x = 0,we see that in this case T = 2p (b/c).

15.For a total swing back and forth of 4.0 cm, the amplitude A is 2.0 cm.

For x(t) = A cos 2pt/T, v(t) = dx/dt = - A2p/T sin 2pt/T.The maximum velocity of the pendulum occurs at the center of theswing equal to 10 cm/s.

|vmax| = 2pA/T. T = 2pA/vmax= 2p (2.0 cm)/10 cm/s = 0.4p s.

16.x(t) = 4.0 cm cos (pt s-1 - p/6).2.0 cm = 4.0 cm cos (pt s-1 - p/6).cos (pt s-1 - p/6) = 0.5 and (pt s-1 - p/6) = p/3 (60o).v(t) = dx/dt = - 4.0p cm/s sin (pt s-1 - p/6).When (pt s-1 - p/6) = p/3, sin p/3 = 0.866 and v = - 4.0p (0.866)cm/s= - 10.9 cm/s.

17.

w= 2pf = 2p (3/2 s-1) = 3p s-1. xo = 0.25 m and vo = - 1.5 m/s.

We know that 0 < d < p/2 because xo is positive and vo is negative.The initial position of the object is less than the amplitude, but

positive, and moving toward the equilibrium position with a negativevelocity.

As shown in #8 above,A = [(xo)2 + (- vo/w)2]1/2 = [(0.25 m)2 + (1.5 m/s/3p s-1)2]1/2 = 0.296 m.tan d = - vo/wxo = 1.5 m/s/(0.25 m)(3p s-1) = 0.637.tan-1 0.637 = 0.18p = 32.5o.x(t) = A cos (2pt/T + d) = 0.296 m cos (3pt s-1 + 0.18p).



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Plots of position x and the velocity v as a function of t are shown in17a and 17b above, respectively. Note xo = 0.25 m and vo = - 1.5m/s.

18.

Fnet = ma = m d2x/dt2

For Fig. 5a,- k1x - k2x = - (k1 + k2)x = m d2x/dt2 ord2x/dt2 + [(k1 + k2)/m]x = 0.Compare withd2x/dt2 + [k/m]x = 0 when f = (1/2p)(k/m)1/2

and find for this case,f = (1/2p)[(k1 + k2)/m]1/2.The "effective" spring constant for springs in parallel is keff= k1 + k2. . + kn.

For Fig. 5b, the spring with constant k2 is in contact with mass m thathas a displacement x = x1 + x2, where x1 is the extension of thespring with constant k1 and x2 is the extension of the spring withconstant k2. The force on the object is:- k2x2 and - k2x2 = m d2x/dt2 (Equation 1)Also,x = x1 + x2 (Equation 2)and the magnitude of the force on the second spring due to the firstspring equals the magnitude of force on the first spring due to thesecond spring, ork1x1 = k2x2 or x1 = k2x2/k1 (Equation 3)Substituting Eq. 3 into Eq. 2,

x = (k2x2/k1) + x2 = (k1 + k2)x2/k1 , orx2 = k1x/(k1 + k2) (Equation 4)Substituting Eq. 4 into Eq. 1,- [k2k1/(k1 + k2)]x = m d2x/dt2 ord2x/dt2 + [{k1k2/(k1 + k2}/m)]x = 0.Comparing withd2x/dt2 + [k/m]x = 0For this case,



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f = (1/2p)[{k1k2/(k1 + k2)}/m]1/2.The "effective" spring constant for series is:keff= {k1k2/(k1 + k2} or1/keff= 1/k1 + 1/k2 + . .+ 1/kn.

19.Imagine the spring cut into thirds with each part having a springconstant k. When the three springs are connected in series, thespring constant isk = 10.0 N/m.For series,1/k = 1/10 N/m = 1/k + 1/k + 1/k= 3/kk = 3k = 30.0 N/m.When two of these springs with k are connected in series,

1/k = 1/k + 1/k = 2/30 N/m,or the spring constant with 2/3 of the spring left (1/3 cut off) isk = 15 N/m.T= 2p(m/k)12 = 2p(0.30/15)1/2 s = 2p (0.02)1/2 s.

20.

a. The differential mass dm = msdy/L(Fig. 6 above)

b. Assuming the velocity vy at y increases linearly with y from 0 aty = 0to v at y = L, vy = vy/L.

c.dK = 1/2 dm vy

2

= 1/2(msdy/L)(vy/L)

2

.

The effective mass that takes place in the oscillation is ms/3.The fraction of the mass is 1/3.



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21.

a. The slope of force F as a function of the extension x is

k = (5.0 - 0)N/(0.50 - 0)m = 10 N/m.

b. For a total mass M and force constant k, the period of themotion is

T = 2p(M/k)1/2.



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For a spring of effective mass ms, M = m + ms, where m isthe variable mass added to the spring load. For this case

T = 2p (m + ms)/k]1/2 orT2 = (4p2/k)m +(4p2/k)ms (Equation 1)

If we compare Eq. 1 to the equation of a line, y = (slope)x +(y-intercept), we see that T2 versus m should yield a straightline and the intercept on the T2 axis is (4p2/k)ms. From Fig. 7aabove, we find k = 10 N/m. From Fig. 7b above, we find theintercept on the T2 axis = 0.154 s2 = (4p2/k)ms = (4p2/10N/m)ms. ms = (1.54/4p2)kg = 0.390 kg and ms = 3 x (0.390kg) = 0.117 kg = 117 g.

c.When T

2

= 0 in Eq. 1, 0 = (4p

2

/k)m + (4p

2

/k)ms orms = - m = - (-0.0390 kg).ms = 0.0390 kg, agreeing with the value found in part (b).

d. P.S. It helps to have my graphing program that reads offcoordinates.

22.

For simple harmonic motion, a/x = - w2.For maximum acceleration, |amax| = Aw2.For motion in the vertical direction, (Fnet)y = may = m(0) orN - mg = 0 and N = mg.

The frictional force that keeps the block from slipping on the plate f= N = mg = ma = mAw2. A = g/w2 = 0.60(10 m/s2)/(1.5 s-1)2 =2.7 m.

23.a.

For the equilibrium position, dU/dr = 0 = (-5a/r

6

) + (3b/r

4

).Calling r = ro at the equilibrium position, ro = (5a/3b)1/2.



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b.

The reduced mass = m/2 and the frequency f = (1/2p)(k/)1/2

=(1/2p)[(12b/m)(3b/5a)5/2]1/2.

24.

a. U = mgy

b. dU/dx = d(mgy)/dx = mg.d[R - (R2 - x2)1/2]/dx = mgx/(R2 - x2)1/2.For x < < R, dU/dx = mgx/R. Fx = - dU/dx = - (mg/R)x.The force Fx is directly proportional to the displacement x andin the opposite direction. The motion is simple harmonic.

c. Fx = ma or- (mg/R)x = m d2x/dt2 and d2x/dt2 + (g/R)x = 0.Compare d2x/dt2 + (k/m)x = 0 (for which T = 2p(m/k)1/2)

and see for this case T= 2p(R/g)1/2.

25.

a. When x = A, v = 0 and K = 0. In general, E = U + K = 1/2 kx2

+ 1/2 mv2. When x = A, E = 1/2 kA2 + 1/2 m(0) = 1/2 kA2.Since E is a constant, E always equals:

1/2 kA2

= 1/2 kx2

+ 1/2 mv2

(Equation 1)

b. Multiplying Eq. 1 by 2/m gives:

(k/m)A2 = (k/m)x2 +1/2 mv2 or(k/m)(A2 - x2) =v2 and



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v =dx/dt =(k/m)1/2(A2 - x2)1/2.

c. Separating variables dx/(A2 - x2)1/2 = (k/m)1/2 dt.

Let x = A sin Q, dx = A cos Q dQ.(A2 - x2)1/2 = A(1 - sin2 Q)1/2 = A cos Q.

For limits on Q, xo = A sin Qo and sin Qo = xo/A.The lower limit is sin -1 xo/A and the upper is sin -1 x/A.

Integrating the above equation gives:

sin -1 x/A + sin -1 xo/A = (k/m)1/2t orsin -1 x/A = (k/m)1/2t + sin -1 xo/A.

Letting sin -1 xo/A = d,

sin -1 x/A = (k/m)1/2t + d orx(t) = A sin [(k/m)1/2t + d].

26.

a. t = r x F. t= rF sin r, F. About the pivot point, the torquefor the rod of mass m1 is -(L/2)m1g sin Q and for the pointmass it is -Lm2g sin Q. The negative signs occur because theyare restoring torques. When the pendulum is swingingcounterclockwise, the torque tends to make it swingclockwise. The moment of inertia of the rod about an end is1/3 m1L2. The moment of inertia of a point particle of mass m2a distance L from the axis is m2L2.

b. t = Ia



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-Lg(m1/2 + m2)sin Q= (m1/3 + m2)L2 d2Q/dt2

For small Q, sin Q is approximately equal to Q and

d2Q/dt2 + [g(m1/2 + m2)/(m1/3 + m2)L]Q = 0.

Comparing with

d2x/dt2 + [k/m]x = 0 (for which Period = 2p (m/k)1/2),

we find for this pendulum,

Period = 2p [2(m1 + 3m2)L/(m1 + 2m2)3g]1/2

27.

For oscillations about pin axis (Fig. 24 above),

a. t = r x F.t = rF sin r,F = -rmg sin Q -rmg Q for small Q.The moment of inertia about the pin a distance r from thecenter of mass,

I = (1/2 mR2 + mr2).

b. t= Ia = (I)d2Q/dt2

-rmgQ = (1/2 mR2 + mr2)d2Q/dt2

c. or

d2Q/dt2 + [rg/(1/2 R2 + r2)]Q = 0.

Compare with



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d2x/dt + [k/m]x = 0 (for which Period = 2p[m/k]1/2)

and see for the disk,

Period = 2p[(R2 + 2r2)/2rg]1/2.

d. For a minimum period,

d(Period)/dr = 0 =2p(1/2)[(2rg)(4r) - (R2 + 2r2)2g]/4r2g2][(R2 + 2r2)/2rg]1/2 or(2rg)(4r) = (R2 + 2r2)2g or4r2 = R2 + 2r2 and 2r2 = R2 orr = R/(2)1/2.

Source:

http://www.wellesley.edu/Physics/phyllisflemingphysics/107_p_harmonic.html

http://www.wellesley.edu/Physics/phyllisflemingphysics/107_p_harmonic.htmlhttp://www.wellesley.edu/Physics/phyllisflemingphysics/107_p_harmonic.html

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